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Minchanka [31]
4 years ago
8

I also need help with this one ?

Mathematics
1 answer:
uranmaximum [27]4 years ago
7 0

Step-by-step explanation:

Line is intersecting x - axis at (-4, 0) & y - axis at (0, 2)

\frac{x-x_1 }{x_1 - x_2 }  =  \frac{y-y_1 }{y_1 - y_2 }  \\  \\  \therefore \:  \frac{x-( - 4) }{ - 4 - 0 }  =  \frac{y-0 }{0 - 2 }  \\  \\ \therefore \:  \frac{x + 4 }{ - 4 }  =  \frac{y}{ - 2 }  \\  \\ \therefore \:  \frac{x + 4 }{ 2 }  =  \frac{y}{ 1 }  \\    \\ \therefore \:y = \frac{x + 4 }{ 2 }  \\    \\  \huge \pink{ \boxed{\therefore \:y = \frac{x}{ 2 }  + 2}} \\

Is the required equation of line.

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Prove that tan60+tan70-tan50+tan10=2sqrt3?
LiRa [457]
{tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)} 

<span>ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to: </span>

<span>= 8*tan(10)/{1 - 3*tan²(10)} </span>

<span>iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10) </span>

<span>= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)} </span>

<span>= 3*tan(30) = 3*(1/√3) = √3 [Proved] </span>

<span>[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)}, </span>
<span>{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]</span>
8 0
4 years ago
Help me please 5points only sry
AlladinOne [14]

x----chickens

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50x+30y=550

44x+36y=532

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44x+36y=532

x=11-3/5y

44(11-3/5y)+36y=532

x=11-3/5y

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x=11-3/5y

2420+48y=2660

x=11-3/5y

48y=2660-2420

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5 0
4 years ago
The Sanchez family uses 2 3/4 tubes of tooth paste each month. How many tubes of toothpaste does the Sanchez family use each yea
Licemer1 [7]
2.75 * 12 = 33.

If the Sanchez family is using 2 3/4 (or 2.75) tubes of toothpaste and there are 12 months in a year, we use the equation above and get 33 as our answer.
8 0
3 years ago
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Write a definite integral that represents the area of the region. (Do not evaluate the integral.)
joja [24]

Answer:

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2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

\int_{a}^{b}|f(x)-g(x)dx

\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx

 

8 0
3 years ago
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